Find the Roots (Zeros) 2x^4-5x^3-17x^2+41x-21=0

$2 x^{4} - 5 x^{3} - 17 x^{2} + 41 x - 21 = 0$

Step 1

Factor the left side of the equation.

Factor $2 x^{4} - 5 x^{3} - 17 x^{2} + 41 x - 21$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 21, \pm 3, \pm 7 q = \pm 1, \pm 2$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.5, \pm 21, \pm 10.5, \pm 3, \pm 1.5, \pm 7, \pm 3.5$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$2 \cdot 1^{4} - 5 \cdot 1^{3} - 17 \cdot 1^{2} + 41 \cdot 1 - 21$

Raise $1$ to the power of $4$ .

$2 \cdot 1 - 5 \cdot 1^{3} - 17 \cdot 1^{2} + 41 \cdot 1 - 21$

Multiply $2$ by $1$ .

$2 - 5 \cdot 1^{3} - 17 \cdot 1^{2} + 41 \cdot 1 - 21$

Raise $1$ to the power of $3$ .

$2 - 5 \cdot 1 - 17 \cdot 1^{2} + 41 \cdot 1 - 21$

Multiply $- 5$ by $1$ .

$2 - 5 - 17 \cdot 1^{2} + 41 \cdot 1 - 21$

Subtract $5$ from $2$ .

$- 3 - 17 \cdot 1^{2} + 41 \cdot 1 - 21$

Raise $1$ to the power of $2$ .

$- 3 - 17 \cdot 1 + 41 \cdot 1 - 21$

Multiply $- 17$ by $1$ .

$- 3 - 17 + 41 \cdot 1 - 21$

Subtract $17$ from $- 3$ .

$- 20 + 41 \cdot 1 - 21$

Multiply $41$ by $1$ .

$- 20 + 41 - 21$

Add $- 20$ and $41$ .

$21 - 21$

Subtract $21$ from $21$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{4} - 5 x^{3} - 17 x^{2} + 41 x - 21}{x - 1}$

Divide $2 x^{4} - 5 x^{3} - 17 x^{2} + 41 x - 21$ by $x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

Divide the highest order term in the dividend $2 x^{4}$ by the highest order term in divisor $x$ .

					$2 x^{3}$
	$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$

Multiply the new quotient term by the divisor.

				$2 x^{3}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			+	$2 x^{4}$	-	$2 x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{4} - 2 x^{3}$

				$2 x^{3}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{3}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{3}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$

Divide the highest order term in the dividend $- 3 x^{3}$ by the highest order term in divisor $x$ .

				$2 x^{3}$	-	$3 x^{2}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$

Multiply the new quotient term by the divisor.

				$2 x^{3}$	-	$3 x^{2}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$
					-	$3 x^{3}$	+	$3 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 x^{3} + 3 x^{2}$

				$2 x^{3}$	-	$3 x^{2}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$
					+	$3 x^{3}$	-	$3 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{3}$	-	$3 x^{2}$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$
					+	$3 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

$2 x^{3}$

$3 x^{2}$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

Divide the highest order term in the dividend $- 20 x^{2}$ by the highest order term in divisor $x$ .

$2 x^{3}$

$3 x^{2}$

$20 x$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

Multiply the new quotient term by the divisor.

$2 x^{3}$

$3 x^{2}$

$20 x$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

$20 x^{2}$

$20 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 20 x^{2} + 20 x$

$2 x^{3}$

$3 x^{2}$

$20 x$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

$20 x^{2}$

$20 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$2 x^{3}$

$3 x^{2}$

$20 x$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

$20 x^{2}$

$20 x$

$21 x$

Pull the next terms from the original dividend down into the current dividend.

$2 x^{3}$

$3 x^{2}$

$20 x$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

$20 x^{2}$

$20 x$

$21 x$

$21$

Divide the highest order term in the dividend $21 x$ by the highest order term in divisor $x$ .

$2 x^{3}$

$3 x^{2}$

$20 x$

$21$

$x$

$1$

$2 x^{4}$

$5 x^{3}$

$17 x^{2}$

$41 x$

$21$

$2 x^{4}$

$2 x^{3}$

$3 x^{3}$

$17 x^{2}$

$3 x^{3}$

$3 x^{2}$

$20 x^{2}$

$41 x$

$20 x^{2}$

$20 x$

$21 x$

$21$

Multiply the new quotient term by the divisor.

				$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$
					+	$3 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$	+	$41 x$
							+	$20 x^{2}$	-	$20 x$
									+	$21 x$	-	$21$
									+	$21 x$	-	$21$

The expression needs to be subtracted from the dividend, so change all the signs in $21 x - 21$

				$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$
					+	$3 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$	+	$41 x$
							+	$20 x^{2}$	-	$20 x$
									+	$21 x$	-	$21$
									-	$21 x$	+	$21$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
$x$	-	$1$		$2 x^{4}$	-	$5 x^{3}$	-	$17 x^{2}$	+	$41 x$	-	$21$
			-	$2 x^{4}$	+	$2 x^{3}$
					-	$3 x^{3}$	-	$17 x^{2}$
					+	$3 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$	+	$41 x$
							+	$20 x^{2}$	-	$20 x$
									+	$21 x$	-	$21$
									-	$21 x$	+	$21$
												$0$

Since the remander is $0$ , the final answer is the quotient.

$2 x^{3} - 3 x^{2} - 20 x + 21$

Write $2 x^{4} - 5 x^{3} - 17 x^{2} + 41 x - 21$ as a set of factors.

$(x - 1) (2 x^{3} - 3 x^{2} - 20 x + 21) = 0$

Factor $2 x^{3} - 3 x^{2} - 20 x + 21$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 21, \pm 3, \pm 7 q = \pm 1, \pm 2$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.5, \pm 21, \pm 10.5, \pm 3, \pm 1.5, \pm 7, \pm 3.5$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$2 \cdot 1^{3} - 3 \cdot 1^{2} - 20 \cdot 1 + 21$

Raise $1$ to the power of $3$ .

$2 \cdot 1 - 3 \cdot 1^{2} - 20 \cdot 1 + 21$

Multiply $2$ by $1$ .

$2 - 3 \cdot 1^{2} - 20 \cdot 1 + 21$

Raise $1$ to the power of $2$ .

$2 - 3 \cdot 1 - 20 \cdot 1 + 21$

Multiply $- 3$ by $1$ .

$2 - 3 - 20 \cdot 1 + 21$

Subtract $3$ from $2$ .

$- 1 - 20 \cdot 1 + 21$

Multiply $- 20$ by $1$ .

$- 1 - 20 + 21$

Subtract $20$ from $- 1$ .

$- 21 + 21$

Add $- 21$ and $21$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} - 3 x^{2} - 20 x + 21}{x - 1}$

Divide $2 x^{3} - 3 x^{2} - 20 x + 21$ by $x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$2 x^{3}$

$3 x^{2}$

$20 x$

$21$

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x$ .

					$2 x^{2}$
	$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			+	$2 x^{3}$	-	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} - 2 x^{2}$

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$

Divide the highest order term in the dividend $- x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$x$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$x$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					-	$x^{2}$	+	$x$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{2} + x$

				$2 x^{2}$	-	$x$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	-	$x$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$
							-	$21 x$

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$	-	$x$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$
							-	$21 x$	+	$21$

Divide the highest order term in the dividend $- 21 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$x$	-	$21$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$
							-	$21 x$	+	$21$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$x$	-	$21$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$
							-	$21 x$	+	$21$
							-	$21 x$	+	$21$

The expression needs to be subtracted from the dividend, so change all the signs in $- 21 x + 21$

				$2 x^{2}$	-	$x$	-	$21$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$
							-	$21 x$	+	$21$
							+	$21 x$	-	$21$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	-	$x$	-	$21$
$x$	-	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$20 x$	+	$21$
			-	$2 x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$20 x$
					+	$x^{2}$	-	$x$
							-	$21 x$	+	$21$
							+	$21 x$	-	$21$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$2 x^{2} - x - 21$

Write $2 x^{3} - 3 x^{2} - 20 x + 21$ as a set of factors.

$(x - 1) ((x - 1) (2 x^{2} - x - 21)) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 21 = - 42$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$(x - 1) ((x - 1) (2 x^{2} - (x) - 21)) = 0$

Rewrite $- 1$ as $6$ plus $- 7$

$(x - 1) ((x - 1) (2 x^{2} + (6 - 7) x - 21)) = 0$

Apply the distributive property.

$(x - 1) ((x - 1) (2 x^{2} + 6 x - 7 x - 21)) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) ((x - 1) ((2 x^{2} + 6 x) - 7 x - 21)) = 0$

Factor out the greatest common factor (GCF) from each group.

$(x - 1) ((x - 1) (2 x (x + 3) - 7 (x + 3))) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x - 1) ((x - 1) ((x + 3) (2 x - 7))) = 0$

Remove unnecessary parentheses.

$(x - 1) ((x - 1) (x + 3) (2 x - 7)) = 0$

Remove unnecessary parentheses.

$(x - 1) (x - 1) (x + 3) (2 x - 7) = 0$

Combine like factors.

Raise $x - 1$ to the power of $1$ .

$(x - 1) (x - 1) (x + 3) (2 x - 7) = 0$

Raise $x - 1$ to the power of $1$ .

$(x - 1) (x - 1) (x + 3) (2 x - 7) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(x - 1)}^{1 + 1} (x + 3) (2 x - 7) = 0$

Add $1$ and $1$ .

${(x - 1)}^{2} (x + 3) (2 x - 7) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x - 1)}^{2} = 0$

$x + 3 = 0$

$2 x - 7 = 0$

Step 3

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

${(x - 1)}^{2} = 0$

Solve ${(x - 1)}^{2} = 0$ for $x$ .

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 4

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Set $2 x - 7$ equal to $0$ and solve for $x$ .

Set $2 x - 7$ equal to $0$ .

$2 x - 7 = 0$

Solve $2 x - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

Step 6

The final solution is all the values that make ${(x - 1)}^{2} (x + 3) (2 x - 7) = 0$ true.

$x = 1, - 3, \frac{7}{2}$

Step 7

Find the Roots (Zeros) 2x^4-5x^3-17x^2+41x-21=0

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